in this version of the \giant swing\, the seat is connected to two cables, one of which is horizontal…

in this version of the \giant swing\, the seat is connected to two cables, one of which is horizontal (figure 1). the seat swings in a horizontal circle at a rate of 21.0 rpm (rev/min). for general problem - solving tips and strategies for this topic, you may want to view a video tutor solution of a conical pendulum. part a if the seat weighs 305 n and a 875 n person is sitting in it, find the tension t_horizontal in the horizontal cable. express your answer with the appropriate units. part b if the seat weighs 305 n and a 875 n person is sitting in it, find the tension t_inclined in the inclined cable. express your answer with the appropriate units.

in this version of the \giant swing\, the seat is connected to two cables, one of which is horizontal (figure 1). the seat swings in a horizontal circle at a rate of 21.0 rpm (rev/min). for general problem - solving tips and strategies for this topic, you may want to view a video tutor solution of a conical pendulum. part a if the seat weighs 305 n and a 875 n person is sitting in it, find the tension t_horizontal in the horizontal cable. express your answer with the appropriate units. part b if the seat weighs 305 n and a 875 n person is sitting in it, find the tension t_inclined in the inclined cable. express your answer with the appropriate units.

Answer

Explanation:

Step1: Calculate total weight

The total weight $W$ of the seat and the person is the sum of their weights. $W = 305\ N+ 875\ N=1180\ N$.

Step2: Analyze vertical - force equilibrium for part B

In the vertical direction for the inclined - cable situation, if the angle of the inclined cable with the vertical is $\theta = 40.0^{\circ}$, and the tension in the inclined cable is $T_{inclined}$, the vertical component of the tension in the inclined cable balances the total weight. So $T_{inclined}\cos\theta=W$. Then $T_{inclined}=\frac{W}{\cos\theta}$. Substituting $W = 1180\ N$ and $\theta = 40.0^{\circ}$, we have $T_{inclined}=\frac{1180\ N}{\cos40.0^{\circ}}\approx\frac{1180\ N}{0.766}\approx1540\ N$.

Step3: Analyze horizontal - force equilibrium for part A

In the horizontal direction, the tension in the horizontal cable $T_{horizontal}$ is related to the tension in the inclined cable. The horizontal component of the tension in the inclined cable is $T_{inclined}\sin\theta$. So $T_{horizontal}=T_{inclined}\sin\theta$. Since $T_{inclined}=\frac{W}{\cos\theta}$, then $T_{horizontal}=W\tan\theta$. Substituting $W = 1180\ N$ and $\theta = 40.0^{\circ}$, we get $T_{horizontal}=1180\ N\times\tan40.0^{\circ}\approx1180\ N\times0.839\approx990\ N$.

Answer:

Part A: $T_{horizontal}=990\ N$ Part B: $T_{inclined}=1540\ N$